Very hard Physics question

In summary, the conversation begins with a question about a difficult physics problem involving a football being kicked at a 50 degree angle and traveling a horizontal distance of 20m. The conversation then shifts to a policy of not providing complete answers and instead offering guidance. The conversation continues with equations and hints being provided to help solve the problem. The original poster mentions being stuck at one point and another user provides a link to a helpful website. The conversation ends with a user providing a solution and a humorous comment about necroposting.
  • #1

mack2629

Very hard Physics question!

This question is very hard, a football is kicked at an 50 degree angle to the horizontal , travels a horizontal distance of 20m before hitting the ground. what is the initial speed?
please tell me how you would do this
 
Physics news on Phys.org
  • #2


Hi Mack, welcome to PF.

Originally posted by mack2629
This question is very hard,

No, it is not.

please tell me how you would do this

Sorry, but I have a policy here: You show us how you start and where you get stuck, and we help you through the rough spots. I have it stuck to the top of the forum: "Read This Before Posting". We don't do yer homework for you here.

This is a basic projectile motion problem. Start by writing down your knowns and unknowns, and--most importantly--read up on projectile motion.
 
  • #3
If you have read the rules of this forum, we are not able to answer your question completely.
But i will give you some clues.
From these three equations (you may not need them all), you can derive an equation that makes a relation between the horizontal distance travelled, the angle of the kick, and the initial speed (if you already know this equation just use it).
After you get the equation, plug in what you know (the distance and angle), and solve for the unknown (the intial speed).

(you will have to apply the equations on the horizontal movement alone, and the vertical movement alone)
S=vi*t+0.5*a*t2
vf2=vi2+2*a*S
vf=a*t+vi

If you are still stuck, tell us and we will help you a little more .
 
  • #4
i know, V1x=Vcos50
Ax=0m/s*
dx=20m
t=?

V1y=Vsin50
Ay=-9.8m/s*
dy=?
t=?

t=20/Vcos50
V1x=20/t
and I am stuckafter this because when i plug my first equation into the second you can't solve it.

by the way I am in grade 11.
 
Last edited by a moderator:
  • #5
Hint: You might want to use Vsin50 and Vcos50 instead of Vy and Vx. (1)g=Vsin50*(t/2) (2)s=Vcos50*t

Do you know, how (1) is formed?
 
Last edited by a moderator:
  • #6
Okz, here are some little tricks.
Since you are omiting the friction with air, then the energy will be reserved.
Since the energy is reserved, the speed of the object at launch will be exactly equal to the speed of the object when it hist back the ground (but opposite in direction).
So ..
vf=a*t+vi
But vf=-vi
-vi=a*t+vi
-2vi=a*t
t=-2vi/a

Now you can use that equation on the vertical movement to get the time of the whole trip.

S=vi*t+0.5*a*t2
Apply this on the horizontal movement, note that that the only unknowns are vi and t.
Subsitute the value of t from the first equation.
Now you have a single equation (quadratice) with a single variable, solve it.

Still stuck ?
 
  • #7
The only formula you can use is the one for range. I'm not sure if you've even seen it yet.
At any rate,
Range = V^2 * sin(2*Theta) / g
 
  • #9
Hi mack2629, I believe that it always helps to first figure out what is happening.
1) No friction -> path of flight is parabola. Launch speed is equal to impact speed (!)
2) Zeroes of parabola are 0 and 20, so y = ax(x-20).
3) 50 degree angle of launch, so dy/dx = tan 50 when x = 0.
4) Calculus: dy/dx = 2ax-20a, so -20a = tan 50.
Now figure out how high is the highest point of that parabola, and what time it takes to fall from that height, and what speed is reached when hitting the ground after having fallen from that height, and use a little common sense and some Pythagoras, and you 've got it. :wink:
 
  • #10


range=(sin(2a)u^2)/g
 
  • #11


matt717 said:
range=(sin(2a)u^2)/g

Well done you just necroposted in a thread that is 8 years, 9 months, 2 weeks, 1 day old.
 

What is the definition of a "Very hard Physics question"?

A "Very hard Physics question" can be defined as a question that requires a deep understanding of complex concepts and theories in physics, and often requires advanced mathematical skills to solve.

Why are "Very hard Physics questions" important to study?

Studying "Very hard Physics questions" allows scientists to push the boundaries of our understanding of the physical world, and can lead to breakthroughs in technology and advancements in various fields such as engineering and medicine.

What are some common strategies for solving "Very hard Physics questions"?

Some common strategies for solving "Very hard Physics questions" include breaking down the problem into smaller, more manageable parts, using mathematical equations and principles, and applying critical thinking and problem-solving skills.

What are some resources available for help with "Very hard Physics questions"?

There are many resources available for help with "Very hard Physics questions", such as textbooks, online forums and communities, peer-reviewed journals, and consulting with experts in the field.

How can I improve my ability to solve "Very hard Physics questions"?

To improve your ability to solve "Very hard Physics questions", it is important to have a strong foundation in fundamental concepts and theories, practice regularly, and seek help and guidance from experienced physicists or educators.

Suggested for: Very hard Physics question

Replies
9
Views
1K
Replies
11
Views
226
Replies
5
Views
225
Replies
13
Views
598
Replies
13
Views
150
Replies
8
Views
838
Replies
5
Views
192
Replies
22
Views
1K
Replies
5
Views
608
Back
Top